西安电子科技大学-数字信号处理-试卷C答案

Answer to “Digital Signal Processing of 2005”

Problem 1

(a) even part: };5.0,1,7,7,5,7,7,1,5.0{---=e X odd part: };5.0,1,3,1,0,1,3,1,5.0{----=o X

(b) };20,16,11,94,36,40,31

,16,12,0{-----=y (c) MATLAB Program

n=-4:2;

x=[1 -2 4 6 -5 8 10]; [x11,n11]=sigshift(x,n,2); [x12,n12]=sigshift(x,n,-1); [x13,n13]=sigfold(x,n); [x13,n13]=sigshift(x13,n13,-2); [x12,n12]=sigmult(x,n,x12,n12); [y,n]=sigadd(2*x11,n11,x12,n12); [y,n]=sigadd(y,n,-1*x13,n13)

Problem 2

(a)w j w j w j w j jw jw e e e e e e X 65424210124)(-----++++++=，()j X e ωis periodic in ω with period 2π

(b) MATLAB Program :

clear; close all;

n = 0:6; x = [4,2,1,0,1,2,4]; w = [0:1:1000]*pi/1000;

X = x*exp(-j*n'*w); magX = abs(X); phaX = angle(X); % Magnitude Response Plot

subplot(2,1,1); plot(w/pi,magX);grid;

xlabel('frequency in pi units'); ylabel('|X|'); title('Magnitude Response'); % Phase response plot

subplot(2,1,2); plot(w/pi,phaX*180/pi);grid;

xlabel('frequency in pi units'); ylabel('Degrees'); title('Phase Response'); axis([0,1,-180,180])

(c) Because the given sequence x (n)={4,2,1,0,1,2,4} (n=0,1,2,3,4,5,6) is symmetric about 1

32

N α-=

=,the phase response ()j H e ω

< satisfied the condition ：()3j H e ω

αωω<=-=- so the phase response is a linear function in ω.

(d) 150,350Hz Hz Ω=-;

(e) The difference of amplitude and magnitude response:

Firstly, the amplitude response is a real function, and it may be both positive and negative. The magnitude response is always positive.

Secondly, the phase response associated with the magnitude response is a discontinuous function. While the associated with the amplitude is a continuous linear function.

Problem 3

(a) )9.09.01/()1()(211------=z z z z H

Zero:0 and 1; Pole:-0.6 and 1.5; (b)11

16151()212110.61 1.5H z z z

--=

⨯+⨯+-， 165()((0.6)(1.5))()2121n n

h n u n =-+ (c) ROC : 0.6 1.5Z <<，

()()()163531215212

n n

h n u n u n ⎛

⎫⎛⎫=⨯--⨯--- ⎪ ⎪⎝⎭⎝⎭ Problem 4

(a) y(n)={50,44,34,52};

(b) y(n)={5,16,34,52,45,28,0}; (c) N=6;

(d) MATLAB Program :

Function y=circonv(x1,x2,N) If (length(x1)>N)

error(“N must not be smaller than the length of sequence ”) else

x1=[x1,zeros(1,N-length(x1))]; end

if(length(x2)>N)

error(“N must not be smaller than the length of sequence ”)

else

x2=[x2,zeros(1,N-length(x2))]; end

y1=dft(x1,N).*dft(x2,N); y=idft(y,N);

(e) DTFT is discrete in time domain, but continuous in frequency domain. The DFT is discrete both in time and frequency domain.The FFT is a very efficient method for calculating DFT.

Problem 5

(a) Direct form II uses the little delay and it can decrease the space of the compute. (b)The advantage of the linear-phase form:

1. For frequency-selective filters, linear-phase structure is generally desirable to have a phase-response

that is a linear function of frequency.

2. This structure requires 50% fewer multiplications than the direct form. (c) Block diagrams are shown as under:

1

z -1

z -1

z -1

z -1

z -1

z -

()

x n )

n

()

x n

1-1

-

Problem 6

(a) we use Hamming or Blackman window to design the bandpass filter because it can provide us attenuation

exceed 60dB .

(b) According to Blackman window :

first, Determine transition width =

p s W W - ;

second, Determine the type of the window according to s A ;

third, Compute M using the formula M

W W p s π2=

- ; fourth, Compute ideal LPF

2

s

p W W Wc +=

;

fifth, design the window needed, multiply point by point; sixth, determine

s p A R ,

(c) MATLAB Program :